【最短路】POJ 1511 Invitation Cards

POJ 1511 Invitation Cards

题意：给一个有向图，求从结点1到各结点，再从各结点回到结点1的最小花费之和。

思路：

反向建边，没了。

一开始INF设成了0x3f3f3f3f，以为真的可以当无穷大用，洛谷那边WA一发才发现它表示的数量级是\(10^9\)，在本题不够用，遂改为1e17。然后TLE，发现关了同步的cin还是不够快，遂抄了个快读函数，过了。

换到POJ这边，一模一样的题，代码复制过去交了，WA。最后把INF改为0x3f3f3f3f3f3f3f3f，过了。奇妙评测机……

const LL INF = 0x3f3f3f3f3f3f3f3f;
const int maxn = 1000000+100;
LL read() {
	LL x = 0, f = 1;char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1;c = getchar();}
	while (c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
	return x * f;
}

struct node {
	LL d;
	int u;
	bool operator < (const node& k) const {
		return d > k.d;
	}
};

struct Edge {
	int from, to;
	LL dis;
	Edge(int u,int v,LL d):from(u),to(v),dis(d){}
};

int n, m,x;
int p[maxn];
vector<Edge> Edges;
vector<Edge> A_Edges;
vector<int> G[maxn];
vector<int> A_G[maxn];
bool done[maxn];
LL d[maxn];
LL A_d[maxn];


void dij(int start, int A) {
	memset(done, 0, sizeof(done));
	for (int i = 1; i <= n; i++){
		if (A == 0) d[i] = INF;
		else A_d[i] = INF;
	}
	if (A == 0) d[start] = 0;
	else A_d[start] = 0;

	priority_queue<node> Q;
	Q.push(node{ 0,start });

	while (!Q.empty()) {
		node x = Q.top(); Q.pop();
		int u = x.u;
		if (done[u]) continue;
		if (A == 0) {
			for (int i = 0; i < G[u].size(); i++) {
				Edge& e = Edges[G[u][i]];
				if (d[e.to] > d[u] + e.dis) {
					d[e.to] = d[u] + e.dis;
					Q.push(node{ d[e.to],e.to });
				}
			}
		}
		else {
			for (int i = 0; i < A_G[u].size(); i++) {
				Edge& e = A_Edges[A_G[u][i]];
				if (A_d[e.to] > A_d[u] + e.dis) {
					A_d[e.to] = A_d[u] + e.dis;
					Q.push(node{ A_d[e.to],e.to });
				}
			}
		}
		done[u] = true;
	}
}

void solve(){
	n = read(); m = read();
	Edges.clear();
	A_Edges.clear();
	for (int i = 0; i <= n; i++) G[i].clear();
	for (int i = 0; i <= n; i++) A_G[i].clear();
	memset(done, 0, sizeof(done));

	Edges.push_back(Edge(0, 0, 0));
	A_Edges.push_back(Edge(0, 0, 0));

	int num = 0;
	for (int i = 1; i <= m; i++) {
		int u, v;
		LL d;
		u = read();
		v = read();
		d = read();
		Edges.push_back(Edge(u, v, d));
		A_Edges.push_back(Edge(v, u, d));
		G[u].push_back(++num);
		A_G[v].push_back(num);
	}
	
	dij(1, 0);
	dij(1, 1);

	LL ans = 0;
	for (int i = 2; i <= n; i++) {
		if (d[i] < INF) ans += d[i];
		if (A_d[i] < INF) ans += A_d[i];
	}
	cout << ans<<endl;
}